Solution Procedure
Considering the spool as a cylindrical shape, and calculating the circumference based on the area cited, we establish the connection between circumference and area as follows:
... C = 2√(πA)
To find ten times this circumference, we calculate:
... 10C = 20√(π·20 cm²) = 40√(5π) cm ≈ 159 cm
Derivation of Formulas
The standard formulas for circumference and area are:
... C = 2πr
... A = πr²
Multiplying the area formula by π and extracting the square root yields:
... πA = (πr)²
... √(πA) = πr
Doubling this value results in the circumference formula:
... C = 2√(πA) = 2πr
Response:
The measure of mHLK is "(204)°".
Step-by-step breakdown:
Given values include:
mJI = (3x+2)°
mHLK = (15x-36)°
and,
m∠HML = (8x-1)°
then,
What is mHLK?
Now,
Utilizing the chord-chord angle formula, we find
Inserting the known values into the equation gives us
⇒ 
By carrying out cross-multiplication, we arrive at
⇒ 
⇒ 
By subtracting "18x" from both sides, we obtain
⇒ 
⇒ 
Upon adding "2" to both sides, we end up with
⇒ 
⇒ 
⇒ 
⇒ 
By substituting the value of "x" into mHLK = (15x-36)°, we calculate
⇒ (15x-36)° = (15×16-36)°
⇒ = (240-36)°
⇒ = (204)°
Thus, mHLK = (204)°
4x+32
4 goes into 4 1 time. (4/4=1)
4 goes into 32 8 times. (32/4=8)
We can factor out 4 from both numbers.
4(1x+8)
aka 4(x+8)
It is necessary for the value of m to exceed that of n. When binomials are multiplied, the middle term emerges from combining the outside and inside products. Thus, bx = –nx + mx, simplifying further leads to b = –n + m. When adding numbers that have opposite signs, we subtract their absolute values and retain the sign of the number with the larger absolute value. Since b is positive, m must indeed possess a greater absolute value.
Response: The following comparison is made.
Detailed explanation:
At the playground, a climbing structure is 6 feet high and stretches 4 feet horizontally.
The slope can be calculated as vertical height divided by horizontal length.
= 6/4 = 3/2 = 1.5
In contrast, a climbing structure in the gym stands 10 feet tall, extending 6 feet horizontally.
Therefore, slope, 
= 10/6 = 5/3 = 1.67 (approximately)
As such, <
Thus, the slope of the first climbing structure is less than that of the second.
